--- title: "Footing for 400x300 column, reversible moment" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-for-400x300-column-reversible-moment" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-for-400x300-column-reversible-moment site: BCT Engineering --- # Footing for 400x300 column, reversible moment Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2079 Baisakh, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-for-400x300-column-reversible-moment - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design a RC footing for a column having x-sectional dimension 400 mm x 300 mm, with 8-16 $\phi$ longitudinal reinforcement, column is subjected to axial compressive load of 1000 kN and reversible bending moment 100 kNm. Consider M20 concrete for footing, M25 concrete for column and Fe415 steel grade for both. Take safe bearing capacity of soil is 200 kN/m$^2$ at a depth at 1.5 m. ## Short answer Provide a 3.6 m × 1.8 m footing, 800 mm deep, with identical top and bottom steel. Along the length, use 16 mm bars @ 150 mm c/c; along the width, central band with 12 bars of 16 mm @ 150 mm c/c and each outer strip with 5 bars of 16 mm @ 180 mm c/c. ## Answer #### Data and assumptions - Column 400 mm × 300 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M25; column bars 8-16 mm. - Service load $P = 1000$ kN, service moment $M = 100$ kN·m. - Allowable (safe) bearing capacity of soil = 200 kN/m² at 1.5 m depth. - Loads are service loads. The moment is reversible, so the same steel is provided on both sides. - Clear cover 50 mm for the footing (IS 456 cl. 26.4.2.2); effective depth $d = D - 60$ mm (cover + half bar). Load factor 1.5. #### Size of footing Weight of footing and backfill is taken as 10% of the column load. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{200} = 5.50\ \text{m}^2 $$ Eccentricity $e = M/P = 100.0/1000.0 = 0.100$ m, which is less than $L/6$, so there is no tension under the base. The size is chosen so that $$ q_{max} = 1.1\frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a $$ Try $L \times B = 3.6 \times 1.8$ m ($A = 6.48$ m²): $q_{max} = 198.0$ kN/m² $\le 200.0$ and $q_{min} = 141.5$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 1000.0 = 1500.0$ kN, $M_u = 150.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 270.1$ kN/m², $p_{u,min} = 192.9$ kN/m² (average 231.5 kN/m²). Self-weight does not cause bending in the footing, so it is excluded. #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3600 - 400)/2 = 1600$ mm; along $B$: $(1800 - 300)/2 = 750$ mm. Along $L$ (critical section at the column face on the side of maximum pressure, linear pressure): $M_{u,L} = B\left[p_f\dfrac{x^2}{2} + (p_{max}-p_f)\dfrac{x^2}{3}\right]$, with $p_f = 235.8$ kN/m² at the face and $x = 1.6$ m: $M_{u,L} = 595.9$ kN·m. Along $B$ (average pressure): $M_{u,B} = 234.4$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 740$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 800$ mm, $d = 740$ mm**, found by the shear checks below. $A_{st} = \dfrac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \dfrac{4.6 M_u}{f_{ck} b d^2}}\right] b d$; minimum steel = 0.12% of $bD$. | Direction | $M_u$ (kN·m) | $A_{st}$ req (mm²) | Provide | $A_{st}$ prov (mm²) | |---|---|---|---|---| | Along $L$ (width $B$) | 595.9 | 2315 | 16 mm @ 150 c/c | 2413 | | Along $B$ (width $L$) | 234.4 | 3456 | 16 mm @ 205 c/c | 3531 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 1800$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{2.00+1} = 0.667$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.667 \times 3456 = 2304$ mm² → 16 mm @ 150 mm c/c (12 bars). - Each outer strip (900 mm wide): remaining steel 576 mm² or minimum 0.12% = 864 mm², whichever is more → 16 mm @ 180 mm c/c (5 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 403.8$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.303$ MPa; $p_t = 0.181\%$, $\tau_c = 0.312$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 8.3$ kN, $\tau_v = 0.003$ MPa; $p_t = 0.133\%$, $\tau_c = 0.288$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+740) + (300+740)] = 4360$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 231.5 \times 1.140 \times 1.040 = 1225.6$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.380$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.92$ MPa (M20, deformed bars): 16 mm bars (both directions) → $L_d = 752$ mm. Available length from column face (less 50 mm end cover): 1550 mm along $L$ (adequate), 700 mm along $B$. Along $B$ the available 700 mm is slightly less than $L_d = 752$ mm, so the bars are given 90° bends at the ends (each bend adds an anchorage value of $8\phi = 128$ mm), which is adequate. #### Bearing stress and transfer of load from column Column area $A_2 = 400\times300 = 0.120$ m². Bearing stress $= P_u/A_2 = 12.50$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 11.25$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 18.00 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 1350 kN $<$ $P_u = 1500$ kN, so the excess 150 kN must be carried by dowels: $A_{s,dow} = \dfrac{150000}{0.87 f_y} = 415$ mm²; the column bars (8-16 mm, 1608 mm²) are continued into the footing as dowels, which is adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 602$ mm (and the same lap with the column bars). Available depth in footing = 800 - 50 - 32 = 718 mm $\ge L_{dc}$, **adequate**. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 800 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.6 m, B = 1.8 m ``` - Bottom mesh: 16 mm @ 150 c/c along $L$ (lower layer); along $B$ (upper layer), 16 mm @ 150 c/c in the central 1.8 m band and 16 mm @ 180 c/c in each outer strip, 50 mm clear cover; dowels 8-16 mm. **Answer:** Footing 3.6 m × 1.8 m × 800 mm thick; 16 mm @ 150 c/c along the length; along the width, 16 mm @ 150 c/c in the central 1.8 m band and 16 mm @ 180 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a reversible moment mean for the footing design? A. The moment can act in either direction, so the same steel is provided on both sides B. The moment is ignored C. The footing carries tension D. Only the bottom face is reinforced Answer: A. Because the moment can change direction, the design is made for the worse case and symmetric reinforcement is used. (Key point: Reversible moment) 2. Column is M25 and footing M20. Which permissible bearing stress governs? A. The footing value 18.00 MPa B. The column value 0.45 x 25 = 11.25 MPa, being the lesser C. 0.45 x 20 = 9.00 MPa D. 0.25 sqrt(25) Answer: B. The lesser of the column concrete value (11.25 MPa) and the footing concrete value (up to 18.00 MPa) governs. (Key point: Governing bearing stress) 3. Bearing capacity is 1350 kN and Pu = 1500 kN. The 150 kN excess is: A. Resisted by increasing the footing size B. Ignored C. Carried by 0.5% minimum dowels only D. Carried by dowels of 150000/(0.87 fy) = 415 mm^2; the 8-16 mm bars supply 1608 mm^2 Answer: D. Excess load divided by 0.87 fy gives the dowel area required, which the column bars exceed. (Key point: Dowels for excess load) 4. Along B, available length is 700 mm but Ld = 752 mm. What does the design do? A. Bars are extended or bent up at the ends with 90 degree bends B. Bar diameter is increased C. Footing depth is reduced D. Nothing, 700 mm is accepted Answer: A. When the straight anchorage is short, bars are extended or given 90 degree bends at the ends to develop their strength. 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